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3 years ago

What is the average mass of fruit purchased in lb

Chemistry

1 answer:

Vera_Pavlovna [14]3 years ago

7 0

Answer:

so first what you would do is the first questions asking what is the average mass of the fruits! so in my head im looking at the graph and go i would add up all the fruits MASS make sure to add the mass only and get the total of the mass that all the fruits have. the seconed question is asking what is the average cost of the first purchase? look at the graph it says the first fruit cost 1.37 dollars. the third and last question is calculate the average cost per mass of the fruit purchassed. so this is a tricky one but ill help you what i think would take the cost of the fruit and divide it by the mass of each fruit! i hope tis helps you can do this! i can also say that question two is trying to make you think over it so dont its litteraly stated to you on the graph lol

Explanation:

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Answer: The pH of resulting solution is 8.7

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For TRIS acid:

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol$

For TRIS base:

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

$0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol$

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of TRIS acid = 8.3

$[\text{TRIS acid}]=\frac{0.005}{0.11}$

$[\text{TRIS base}]=\frac{0.012}{0.11}$

pH = ?

Putting values in above equation, we get:

$pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7$

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